public class Solution72 {
    /**
     * 给定两个单词 word1 和 word2，计算出将 word1 转换成 word2 所使用的最少操作数 。
     * <p>
     * 你可以对一个单词进行如下三种操作：
     * <p>
     * 插入一个字符
     * 删除一个字符
     * 替换一个字符
     * 示例 1:
     * <p>
     * 输入: word1 = "horse", word2 = "ros"
     * 输出: 3
     * 解释:
     * horse -> rorse (将 'h' 替换为 'r')
     * rorse -> rose (删除 'r')
     * rose -> ros (删除 'e')
     * <p>
     * # r o s
     * # 0 1 2 3
     * h 1 1 2 3
     * o 2 2 1 2
     * r 3 2 2 2
     * s 4 3 3 2
     * e 5 4 4 3
     */

    public int minDistance(String word1, String word2) {
        char[] chars1 = word1.toCharArray();
        char[] chars2 = word2.toCharArray();
        int n = word1.length();
        int m = word2.length();

        // if one of the strings is empty
        if (n * m == 0)
            return n + m;

        // array to store the convertion history
        int[][] d = new int[n + 1][m + 1];

        // init boundaries
        for (int i = 0; i < n + 1; i++) {
            d[i][0] = i;
        }
        for (int j = 0; j < m + 1; j++) {
            d[0][j] = j;
        }

        // DP compute
        for (int i = 1; i < n + 1; i++) {
            for (int j = 1; j < m + 1; j++) {
                int left = d[i - 1][j] + 1;
                int down = d[i][j - 1] + 1;
                int left_down = d[i - 1][j - 1];
                if (chars1[i - 1] != chars2[j - 1])
                    left_down += 1;
                d[i][j] = Math.min(left, Math.min(down, left_down));

            }
        }
        return d[n][m];
    }
}
